Products of affine algebraic sets have tensor-product coordinate rings

Equations for $X$ in the first $m$ variables and for $Y$ in the last $n$ variables cut out exactly $X\times Y$. Multiplying polynomial functions in separate variables gives the displayed algebra map. Every polynomial in the $m+n$ coordinates is a sum of products of such polynomials, so this map is surjective.

For injectivity write a kernel element as ${\sum }_{i=1}^r{f}_i\otimes h_i$ with the $h_i$ linearly independent, by eliminating redundant terms in a finite expression. At each $x$, the polynomial function ${\sum }_i{f}_i(x)h_i$ on $Y$ is zero. Linear independence in the function space therefore forces every $f_i(x)=0$. By F1 all $f_i$ are zero. If a factor is empty its coordinate ring and that of the product are zero, so the same conclusion holds. Iteration proves the three-factor formula. Only finite expressions and finite-dimensional elimination were used.